Is this defined by the language? Is there a defined maximum? Is it different in different browsers?
JavaScript has two number types: Number and BigInt.
The most frequently-used number type, Number, is a 64-bit floating point IEEE 754 number.
The largest exact integral value of this type is Number.MAX_SAFE_INTEGER, which is:
253-1, or
+/- 9,007,199,254,740,991, or
nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-one
To put this in perspective: one quadrillion bytes is a petabyte (or one thousand terabytes).
"Safe" in this context refers to the ability to represent integers exactly and to correctly compare them.
From the spec:
Note that all the positive and negative integers whose magnitude is no
greater than 253 are representable in the Number type (indeed, the
integer 0 has two representations, +0 and -0).
To safely use integers larger than this, you need to use BigInt, which has no upper bound.
Note that the bitwise operators and shift operators operate on 32-bit integers, so in that case, the max safe integer is 231-1, or 2,147,483,647.
const log = console.log
var x = 9007199254740992
var y = -x
log(x == x + 1) // true !
log(y == y - 1) // also true !
// Arithmetic operators work, but bitwise/shifts only operate on int32:
log(x / 2) // 4503599627370496
log(x >> 1) // 0
log(x | 1) // 1
Technical note on the subject of the number 9,007,199,254,740,992: There is an exact IEEE-754 representation of this value, and you can assign and read this value from a variable, so for very carefully chosen applications in the domain of integers less than or equal to this value, you could treat this as a maximum value.
In the general case, you must treat this IEEE-754 value as inexact, because it is ambiguous whether it is encoding the logical value 9,007,199,254,740,992 or 9,007,199,254,740,993.
>= ES6:
Number.MIN_SAFE_INTEGER;
Number.MAX_SAFE_INTEGER;
<= ES5
From the reference:
Number.MAX_VALUE;
Number.MIN_VALUE;
console.log('MIN_VALUE', Number.MIN_VALUE);
console.log('MAX_VALUE', Number.MAX_VALUE);
console.log('MIN_SAFE_INTEGER', Number.MIN_SAFE_INTEGER); //ES6
console.log('MAX_SAFE_INTEGER', Number.MAX_SAFE_INTEGER); //ES6
It is 253 == 9 007 199 254 740 992. This is because Numbers are stored as floating-point in a 52-bit mantissa.
The min value is -253.
This makes some fun things happening
Math.pow(2, 53) == Math.pow(2, 53) + 1
>> true
And can also be dangerous :)
var MAX_INT = Math.pow(2, 53); // 9 007 199 254 740 992
for (var i = MAX_INT; i < MAX_INT + 2; ++i) {
// infinite loop
}
Further reading: http://blog.vjeux.com/2010/javascript/javascript-max_int-number-limits.html
In JavaScript, there is a number called Infinity.
Examples:
(Infinity>100)
=> true
// Also worth noting
Infinity - 1 == Infinity
=> true
Math.pow(2,1024) === Infinity
=> true
This may be sufficient for some questions regarding this topic.
Jimmy's answer correctly represents the continuous JavaScript integer spectrum as -9007199254740992 to 9007199254740992 inclusive (sorry 9007199254740993, you might think you are 9007199254740993, but you are wrong!
Demonstration below or in jsfiddle).
console.log(9007199254740993);
However, there is no answer that finds/proves this programatically (other than the one CoolAJ86 alluded to in his answer that would finish in 28.56 years ;), so here's a slightly more efficient way to do that (to be precise, it's more efficient by about 28.559999999968312 years :), along with a test fiddle:
/**
* Checks if adding/subtracting one to/from a number yields the correct result.
*
* #param number The number to test
* #return true if you can add/subtract 1, false otherwise.
*/
var canAddSubtractOneFromNumber = function(number) {
var numMinusOne = number - 1;
var numPlusOne = number + 1;
return ((number - numMinusOne) === 1) && ((number - numPlusOne) === -1);
}
//Find the highest number
var highestNumber = 3; //Start with an integer 1 or higher
//Get a number higher than the valid integer range
while (canAddSubtractOneFromNumber(highestNumber)) {
highestNumber *= 2;
}
//Find the lowest number you can't add/subtract 1 from
var numToSubtract = highestNumber / 4;
while (numToSubtract >= 1) {
while (!canAddSubtractOneFromNumber(highestNumber - numToSubtract)) {
highestNumber = highestNumber - numToSubtract;
}
numToSubtract /= 2;
}
//And there was much rejoicing. Yay.
console.log('HighestNumber = ' + highestNumber);
Many earlier answers have shown 9007199254740992 === 9007199254740992 + 1 is true to verify that 9,007,199,254,740,991 is the maximum and safe integer.
But what if we keep doing accumulation:
input: 9007199254740992 + 1 output: 9007199254740992 // expected: 9007199254740993
input: 9007199254740992 + 2 output: 9007199254740994 // expected: 9007199254740994
input: 9007199254740992 + 3 output: 9007199254740996 // expected: 9007199254740995
input: 9007199254740992 + 4 output: 9007199254740996 // expected: 9007199254740996
We can see that among numbers greater than 9,007,199,254,740,992, only even numbers are representable.
It's an entry to explain how the double-precision 64-bit binary format works. Let's see how 9,007,199,254,740,992 be held (represented) by using this binary format.
Using a brief version to demonstrate it from 4,503,599,627,370,496:
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
|-- 52 bits --| |exponent part| |-- 52 bits --|
On the left side of the arrow, we have bit value 1, and an adjacent radix point. By consuming the exponent part on the left, the radix point is moved 52 steps to the right. The radix point ends up at the end, and we get 4503599627370496 in pure binary.
Now let's keep incrementing the fraction part with 1 until all the bits are set to 1, which equals 9,007,199,254,740,991 in decimal.
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
(+1)
1 . 0000 ---- 0001 * 2^52 => 1 0000 ---- 0001.
(+1)
1 . 0000 ---- 0010 * 2^52 => 1 0000 ---- 0010.
(+1)
.
.
.
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
Because the 64-bit double-precision format strictly allots 52 bits for the fraction part, no more bits are available if we add another 1, so what we can do is setting all bits back to 0, and manipulate the exponent part:
┏━━▶ This bit is implicit and persistent.
┃
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
|-- 52 bits --| |-- 52 bits --|
(+1)
1 . 0000 ---- 0000 * 2^52 * 2 => 1 0000 ---- 0000. * 2
|-- 52 bits --| |-- 52 bits --|
(By consuming the 2^52, radix
point has no way to go, but
there is still one 2 left in
exponent part)
=> 1 . 0000 ---- 0000 * 2^53
|-- 52 bits --|
Now we get the 9,007,199,254,740,992, and for the numbers greater than it, the format can only handle increments of 2 because every increment of 1 on the fraction part ends up being multiplied by the left 2 in the exponent part. That's why double-precision 64-bit binary format cannot hold odd numbers when the number is greater than 9,007,199,254,740,992:
(consume 2^52 to move radix point to the end)
1 . 0000 ---- 0001 * 2^53 => 1 0000 ---- 0001. * 2
|-- 52 bits --| |-- 52 bits --|
Following this pattern, when the number gets greater than 9,007,199,254,740,992 * 2 = 18,014,398,509,481,984 only 4 times the fraction can be held:
input: 18014398509481984 + 1 output: 18014398509481984 // expected: 18014398509481985
input: 18014398509481984 + 2 output: 18014398509481984 // expected: 18014398509481986
input: 18014398509481984 + 3 output: 18014398509481984 // expected: 18014398509481987
input: 18014398509481984 + 4 output: 18014398509481988 // expected: 18014398509481988
How about numbers between [ 2 251 799 813 685 248, 4 503 599 627 370 496 )?
1 . 0000 ---- 0001 * 2^51 => 1 0000 ---- 000.1
|-- 52 bits --| |-- 52 bits --|
The value 0.1 in binary is exactly 2^-1 (=1/2) (=0.5)
So when the number is less than 4,503,599,627,370,496 (2^52), there is one bit available to represent the 1/2 times of the integer:
input: 4503599627370495.5 output: 4503599627370495.5
input: 4503599627370495.75 output: 4503599627370495.5
Less than 2,251,799,813,685,248 (2^51)
input: 2251799813685246.75 output: 2251799813685246.8 // expected: 2251799813685246.75
input: 2251799813685246.25 output: 2251799813685246.2 // expected: 2251799813685246.25
input: 2251799813685246.5 output: 2251799813685246.5
/**
Please note that if you try this yourself and, say, log
these numbers to the console, they will get rounded. JavaScript
rounds if the number of digits exceed 17. The value
is internally held correctly:
*/
input: 2251799813685246.25.toString(2)
output: "111111111111111111111111111111111111111111111111110.01"
input: 2251799813685246.75.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
input: 2251799813685246.78.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
And what is the available range of exponent part? 11 bits allotted for it by the format.
From Wikipedia (for more details, go there)
So to make the exponent part be 2^52, we exactly need to set e = 1075.
To be safe
var MAX_INT = 4294967295;
Reasoning
I thought I'd be clever and find the value at which x + 1 === x with a more pragmatic approach.
My machine can only count 10 million per second or so... so I'll post back with the definitive answer in 28.56 years.
If you can't wait that long, I'm willing to bet that
Most of your loops don't run for 28.56 years
9007199254740992 === Math.pow(2, 53) + 1 is proof enough
You should stick to 4294967295 which is Math.pow(2,32) - 1 as to avoid expected issues with bit-shifting
Finding x + 1 === x:
(function () {
"use strict";
var x = 0
, start = new Date().valueOf()
;
while (x + 1 != x) {
if (!(x % 10000000)) {
console.log(x);
}
x += 1
}
console.log(x, new Date().valueOf() - start);
}());
The short answer is “it depends.”
If you’re using bitwise operators anywhere (or if you’re referring to the length of an Array), the ranges are:
Unsigned: 0…(-1>>>0)
Signed: (-(-1>>>1)-1)…(-1>>>1)
(It so happens that the bitwise operators and the maximum length of an array are restricted to 32-bit integers.)
If you’re not using bitwise operators or working with array lengths:
Signed: (-Math.pow(2,53))…(+Math.pow(2,53))
These limitations are imposed by the internal representation of the “Number” type, which generally corresponds to IEEE 754 double-precision floating-point representation. (Note that unlike typical signed integers, the magnitude of the negative limit is the same as the magnitude of the positive limit, due to characteristics of the internal representation, which actually includes a negative 0!)
ECMAScript 6:
Number.MAX_SAFE_INTEGER = Math.pow(2, 53)-1;
Number.MIN_SAFE_INTEGER = -Number.MAX_SAFE_INTEGER;
Other may have already given the generic answer, but I thought it would be a good idea to give a fast way of determining it :
for (var x = 2; x + 1 !== x; x *= 2);
console.log(x);
Which gives me 9007199254740992 within less than a millisecond in Chrome 30.
It will test powers of 2 to find which one, when 'added' 1, equals himself.
Anything you want to use for bitwise operations must be between 0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 - 1).
The console will tell you that 0x80000000 equals +2147483648, but 0x80000000 & 0x80000000 equals -2147483648.
JavaScript has received a new data type in ECMAScript 2020: BigInt. It introduced numerical literals having an "n" suffix and allows for arbitrary precision:
var a = 123456789012345678901012345678901n;
Precision will still be lost, of course, when such big integer is (maybe unintentionally) coerced to a number data type.
And, obviously, there will always be precision limitations due to finite memory, and a cost in terms of time in order to allocate the necessary memory and to perform arithmetic on such large numbers.
For instance, the generation of a number with a hundred thousand decimal digits, will take a noticeable delay before completion:
console.log(BigInt("1".padEnd(100000,"0")) + 1n)
...but it works.
Try:
maxInt = -1 >>> 1
In Firefox 3.6 it's 2^31 - 1.
I did a simple test with a formula, X-(X+1)=-1, and the largest value of X I can get to work on Safari, Opera and Firefox (tested on OS X) is 9e15. Here is the code I used for testing:
javascript: alert(9e15-(9e15+1));
I write it like this:
var max_int = 0x20000000000000;
var min_int = -0x20000000000000;
(max_int + 1) === 0x20000000000000; //true
(max_int - 1) < 0x20000000000000; //true
Same for int32
var max_int32 = 0x80000000;
var min_int32 = -0x80000000;
Let's get to the sources
Description
The MAX_SAFE_INTEGER constant has a value of 9007199254740991 (9,007,199,254,740,991 or ~9 quadrillion). The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent numbers between -(2^53 - 1) and 2^53 - 1.
Safe in this context refers to the ability to represent integers exactly and to correctly compare them. For example, Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2 will evaluate to true, which is mathematically incorrect. See Number.isSafeInteger() for more information.
Because MAX_SAFE_INTEGER is a static property of Number, you always use it as Number.MAX_SAFE_INTEGER, rather than as a property of a Number object you created.
Browser compatibility
In JavaScript the representation of numbers is 2^53 - 1.
However, Bitwise operation are calculated on 32 bits ( 4 bytes ), meaning if you exceed 32bits shifts you will start loosing bits.
In the Google Chrome built-in javascript, you can go to approximately 2^1024 before the number is called infinity.
Scato wrotes:
anything you want to use for bitwise operations must be between
0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 -
1).
the console will tell you that 0x80000000 equals +2147483648, but
0x80000000 & 0x80000000 equals -2147483648
Hex-Decimals are unsigned positive values, so 0x80000000 = 2147483648 - thats mathematically correct. If you want to make it a signed value you have to right shift: 0x80000000 >> 0 = -2147483648. You can write 1 << 31 instead, too.
Firefox 3 doesn't seem to have a problem with huge numbers.
1e+200 * 1e+100 will calculate fine to 1e+300.
Safari seem to have no problem with it as well. (For the record, this is on a Mac if anyone else decides to test this.)
Unless I lost my brain at this time of day, this is way bigger than a 64-bit integer.
Node.js and Google Chrome seem to both be using 1024 bit floating point values so:
Number.MAX_VALUE = 1.7976931348623157e+308
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Challenge
The shortest program by character count that accepts standard input of the form X-Y R, with the following guarantees:
R is a non-negative decimal number less than or equal to 8
X and Y are non-negative angles given in decimal as multiples of 45° (0, 45, 90, 135, etc.)
X is less than Y
Y is not 360 if X is 0
And produces on standard output an ASCII "arc" from the starting angle X to the ending angle Y of radius R, where:
The vertex of the arc is represented by o
Angles of 0 and 180 are represented by -
Angles of 45 and 225 are represented by /
Angles of 90 and 270 are represented by |
Angles of 135 and 315 are represented by \
The polygonal area enclosed by the two lines is filled with a non-whitespace character.
The program is not required to produce meaningful output if given invalid input. Solutions in any language are allowed, except of course a language written specifically for this challenge, or one that makes unfair use of an external utility. Extraneous horizontal and vertical whitespace is allowed in the output provided that the format of the output remains correct.
Happy golfing!
Numerous Examples
Input:
0-45 8
Output:
/
/x
/xx
/xxx
/xxxx
/xxxxx
/xxxxxx
/xxxxxxx
o--------
Input:
0-135 4
Output:
\xxxxxxxx
\xxxxxxx
\xxxxxx
\xxxxx
o----
Input:
180-360 2
Output:
--o--
xxxxx
xxxxx
Input:
45-90 0
Output:
o
Input:
0-315 2
Output:
xxxxx
xxxxx
xxo--
xxx\
xxxx\
Perl, 235 211 225 211 207 196 179 177 175 168 160 156 146 chars
<>=~/-\d+/;for$y(#a=-$'..$'){print+(map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a),$/}
Perl using say feature, 161 149 139 chars
$ echo -n '<>=~/-\d+/;for$y(#a=-$'"'"'..$'"'"'){say map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}' | wc -c
139
$ perl -E '<>=~/-\d+/;for$y(#a=-$'"'"'..$'"'"'){say map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}'
Perl without trailing newline, 153 143 chars
<>=~/-\d+/;for$y(#a=-$'..$'){print$/,map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}
Original version commented:
$_=<>;m/(\d+)-(\d+) (\d+)/;$e=$1/45;$f=$2/45; # parse angles and radius, angles are 0-8
for$y(-$3..$3){ # loop for each row and col
for$x(-$3..$3){
$t=atan2(-$y,$x)/atan2 1,1; # angle of this point
$t+=8if($t<0); # normalize negative angles
#w=split//,"-/|\\"x2; # array of ASCII symbols for enclosing lines
$s.=!$x&&!$y?"o":$t==$e||$t==$f?$w[$t]:$t>$e&&$t<$f?"x":$";
# if it's origin -> "o", if it's enclosing line, get symbol from array
# if it's between enclosing angles "x", otherwise space
}
$s.=$/;
}
print$s;
EDIT 1: Inlined sub, relational and equality operators return 0 or 1.
EDIT 2: Added version with comments.
EDIT 3: Fixed enclosing line at 360º. Char count increased significantly.
EDIT 4: Added a shorter version, bending the rules.
EDIT 5: Smarter fix for the 360º enclosing line. Also, use a number as fill. Both things were obvious. Meh, I should sleep more :/
EDIT 6: Removed unneeded m from match operator. Removed some semicolons.
EDIT 7: Smarter regexp. Under 200 chars!
EDIT 8: Lots of small improvements:
Inner for loop -> map (1 char)
symbol array from split string -> qw (3 chars)
inlined symbol array (6 chars, together with the previous improvement 9 chars!)
Logical or -> bitwise or (1 char)
Regexp improvement (1 char)
Use arithmethic for testing negative angles, inspired by Jacob's answer (5 chars)
EDIT 9: A little reordering in the conditional operators saves 2 chars.
EDIT 10: Use barewords for characters.
EDIT 11: Moved print inside of loop, inspired by Lowjacker's answer.
EDIT 12: Added version using say.
EDIT 13: Reuse angles characters for fill character, as Gwell's answer does. Output isn't as nice as Gwell's though, that would require 5 additional chars :) Also, .. operator doen't need parentheses.
EDIT 14: Apply regex directly to <>. Assign range operator to a variable, as per Adrian's suggestion to bta's answer. Add version without the final newline. Updated say version.
EDIT 15: More inlining. map{block}#a -> map expr,#a.
Lua, 259 characters
Slightly abuses the non-whitespace character clause to produce a dazzling display and more importantly save strokes.
m=math i=io.read():gmatch("%d+")a=i()/45 b=i()/45 r=i()for y=r,-r,-1 do for x=-r,r do c=m.atan2(y,x)/m.pi*4 c=c<0 and c+8 or c k=1+m.modf(c+.5)io.write(x==0 and y==0 and'o'or c>=a and c<=b and('-/|\\-/|\\-'):sub(k,k)or c==0 and b==8 and'-'or' ')end print()end
Input: 45-360 4
\\\|||///
\\\|||//
\\\\|//
--\\|/
----o----
--//|\\--
////|\\\\
///|||\\\
///|||\\\
Able to handle odd angles
Input: 15-75 8
|/////
|//////
|//////
|//////
///////
|//////-
////---
//-
o
MATLAB, 188 chars :)
input '';[w x r]=strread(ans,'%d-%d%d');l='-/|\-/|\-';[X Y]=meshgrid(-r:r);T=atan2(-Y,X)/pi*180;T=T+(T<=0)*360;T(T>w&T<x)=-42;T(T==w)=-l(1+w/45);T(T==x)=-l(1+x/45);T(r+1,r+1)=-'o';char(-T)
Commented code:
%%# Get the string variable (enclose in quotes, e.g. '45-315 4')
input ''
%%# Extract angles and length
[w x r]=strread(ans,'%d-%d%d');
%%# Store characters
l='-/|\-/|\-';
%%# Create the grid
[X Y]=meshgrid(-r:r);
%%# Compute the angles in degrees
T=atan2(-Y,X)/pi*180;
%%# Get all the angles
T=T+(T<=0)*360;
%# Negative numbers indicate valid characters
%%# Add the characters
T(T>w&T<x)=-42;
T(T==w)=-l(1+w/45);
T(T==x)=-l(1+x/45);
%%# Add the origin
T(r+1,r+1)=-'o';
%%# Display
char(-T)
Mathematica 100 Chars
Out of competition because graphics are too perfect :)
f[x_-y_ z_]:=Graphics#Table[
{EdgeForm#Red,Disk[{0,0},r,{x °,y °}],{r,z,1,-1}]
SetAttributes[f,HoldAll]
Invoke with
f[30-70 5]
Result
alt text http://a.imageshack.us/img80/4294/angulosgolf.png
alt text http://a.imageshack.us/img59/7892/angulos2.png
Note
The
SetAttributes[f, HoldAll];
is needed because the input
f[a-b c]
is otherwise interpreted as
f[(a-b*c)]
GNU BC, 339 chars
Gnu bc because of read(), else and logical operators.
scale=A
a=read()/45
b=read()/45
c=read()
for(y=c;y>=-c;y--){for(x=-c;x<=c;x++){if(x==0)if(y<0)t=-2else t=2else if(x>0)t=a(y/x)/a(1)else if(y<0)t=a(y/x)/a(1)-4else t=a(y/x)/a(1)+4
if(y<0)t+=8
if(x||y)if(t==a||t==b||t==b-8){scale=0;u=(t%4);scale=A;if(u==0)"-";if(u==1)"/";if(u==2)"|";if(u==3)"\"}else if(t>a&&t<b)"x"else" "else"o"};"
"}
quit
MATLAB 7.8.0 (R2009a) - 168 163 162 characters
Starting from Jacob's answer and inspired by gwell's use of any non-whitespace character to fill the arc, I managed the following solution:
[w x r]=strread(input('','s'),'%d-%d%d');
l='o -/|\-/|\-';
X=meshgrid(-r:r);
T=atan2(-X',X)*180/pi;
T=T+(T<=-~w)*360;
T(T>x|T<w)=-1;
T(r+1,r+1)=-90;
disp(l(fix(3+T/45)))
And some test output:
>> arc
0-135 4
\||||////
\|||///-
\||//--
\|/---
o----
I could reduce it further to 156 characters by removing the call to disp, but this would add an extra ans = preceding the output (which might violate the output formatting rules).
Even still, I feel like there are some ways to reduce this further. ;)
Ruby, 292 276 186 chars
x,y,r=gets.scan(/\d+/).map{|z|z.to_i};s=(-r..r);s.each{|a|s.each{|b|g=Math::atan2(-a,b)/Math::PI*180/1%360;print a|b==0?'o':g==x||g==y%360?'-/|\\'[g/45%4].chr: (x..y)===g ?'*':' '};puts}
Nicer-formatted version:
x, y, r = gets.scan(/\d+/).map{|z| z.to_i}
s = (-r..r)
s.each {|a|
s.each {|b|
g = (((Math::atan2(-a,b) / Math::PI) * 180) / 1) % 360
print ((a | b) == 0) ? 'o' :
(g == x || g == (y % 360)) ? '-/|\\'[(g / 45) % 4].chr :
((x..y) === g) ? '*' : ' '
}
puts
}
I'm sure someone out there who got more sleep than I did can condense this more...
Edit 1: Switched if statements in inner loop to nested ? : operator
Edit 2: Stored range to intermediate variable (thanks Adrian), used stdin instead of CLI params (thanks for the clarification Jon), eliminated array in favor of direct output, fixed bug where an ending angle of 360 wouldn't display a line, removed some un-needed parentheses, used division for rounding instead of .round, used modulo instead of conditional add
Ruby, 168 characters
Requires Ruby 1.9 to work
s,e,r=gets.scan(/\d+/).map &:to_i;s/=45;e/=45;G=-r..r;G.map{|y|G.map{|x|a=Math.atan2(-y,x)/Math::PI*4%8;print x|y!=0?a==s||a==e%8?'-/|\\'[a%4]:a<s||a>e ?' ':8:?o};puts}
Readable version:
start, _end, radius = gets.scan(/\d+/).map &:to_i
start /= 45
_end /= 45
(-radius..radius).each {|y|
(-radius..radius).each {|x|
angle = Math.atan2(-y, x)/Math::PI * 4 % 8
print x|y != 0 ? angle==start || angle==_end%8 ? '-/|\\'[angle%4] : angle<start || angle>_end ? ' ' : 8 : ?o
}
puts
}
Perl - 388 characters
Since it wouldn't be fair to pose a challenge I couldn't solve myself, here's a solution that uses string substitution instead of trigonometric functions, and making heavy use of your friendly neighbourhood Perl's ability to treat barewords as strings. It's necessarily a little long, but perhaps interesting for the sake of uniqueness:
($x,$y,$r)=split/\D/,<>;for(0..$r-1){$t=$r-1-$_;
$a.=L x$_.D.K x$t.C.J x$t.B.I x$_."\n";
$b.=M x$t.F.N x$_.G.O x$_.H.P x$t."\n"}
$_=$a.E x$r.o.A x$r."\n".$b;$x/=45;$y/=45;$S=' ';
sub A{$v=$_[0];$x==$v||$y==$v?$_[1]:$x<$v&&$y>$v?x:$S}
sub B{$x<=$_[0]&&$y>$_[0]?x:$S}
#a=!$x||$y==8?'-':$S;
push#a,map{A$_,'\\'.qw(- / | \\)[$_%4]}1..7;
push#a,!$x?x:$S,map{B$_}1..7;
eval"y/A-P/".(join'',#a)."/";print
All newlines are optional. It's fairly straightforward:
Grab user input.
Build the top ($a) and bottom ($b) parts of the pattern.
Build the complete pattern ($_).
Define a sub A to get the fill character for an angle.
Define a sub B to get the fill character for a region.
Build an array (#a) of substitution characters using A and B.
Perform the substitution and print the results.
The generated format looks like this, for R = 4:
DKKKCJJJB
LDKKCJJBI
LLDKCJBII
LLLDCBIII
EEEEoAAAA
MMMFGHPPP
MMFNGOHPP
MFNNGOOHP
FNNNGOOOH
Where A-H denote angles and I-P denote regions.
(Admittedly, this could probably be golfed further. The operations on #a gave me incorrect output when written as one list, presumably having something to do with how map plays with $_.)
C# - 325 319 chars
using System;class P{static void Main(){var s=Console.ReadLine().Split(' ');
var d=s[0].Split('-');int l=s[1][0]-48,x,y,r,a=int.Parse(d[0]),b=int.Parse(d[1]);
for(y=l;y>=-l;y--)for(x=-l;x<=l;)Console.Write((x==0&&y==0?'o':a<=(r=((int)
(Math.Atan2(y,x)*57.3)+360)%360)&&r<b||r==b%360?
#"-/|\"[r/45%4]:' ')+(x++==l?"\n":""));}}
Newlines not significant.
Sample input/output
45-180 8
\||||||||////////
\\|||||||///////
\\\||||||//////
\\\\|||||/////
\\\\\||||////
\\\\\\|||///
\\\\\\\||//
\\\\\\\\|/
--------o
135-360 5
\
\\
\\\
\\\\
\\\\\
-----o-----
----/|\\\\\
---//||\\\\
--///|||\\\
-////||||\\
/////|||||\
Java - 304 chars
class A{public static void main(String[]a){String[]b=a[0].split("-");int e=new Integer(b[1]),r=new Integer(a[1]),g,x,y=r;for(;y>=-r;y--)for(x=-r;x<=r;)System.out.print((x==0&y==0?'o':new Integer(b[0])<=(g=((int)(Math.atan2(y,x)*57.3)+360)%360)&g<e|g==e%360?"-/|\\".charAt(g/45%4):' ')+(x++<r?"":"\n"));}}
More readable version:
class A{
public static void main(String[]a){
String[]b=a[0].split("-");
int e=new Integer(b[1]),r=new Integer(a[1]),g,x,y=r;
for(;y>=-r;y--)for(x=-r;x<=r;)System.out.print((
x==0&y==0
?'o'
:new Integer(b[0])<=(g=((int)(Math.atan2(y,x)*57.3)+360)%360)&g<e|g==e%360
?"-/|\\".charAt(g/45%4)
:' '
)+(x++<r?"":"\n"));
}
}
C (902 byte)
This doesn't use trigonometric functions (like the original perl version), so it's quite ``bloated''. Anyway, here is my first code-golf submission:
#define V(r) (4*r*r+6*r+3)
#define F for(i=0;i<r;i++)
#define C ;break;case
#define U p-=2*r+2,
#define D p+=2*r+2,
#define R *++p=
#define L *--p=
#define H *p='|';
#define E else if
#define G(a) for(j=0;j<V(r)-1;j++)if(f[j]==i+'0')f[j]=a;
#define O(i) for(i=0;i<2*r+1;i++){
main(int i,char**v){char*p,f[V(8)];
int j,m,e,s,x,y,r;p=*++v;x=atoi(p);while(*p!=45)p++;
char*h="0123";y=atoi(p+1);r=atoi(*++v);
for(p=f+2*r+1;p<f+V(r);p+=2*r+2)*p=10;
*(p-2*r-2)=0;x=x?x/45:x;y/=45;s=0;e=2*r;m=r;p=f;O(i)O(j)
if(j>e)*p=h[0];E(j>m)*p=h[1];E(j>s)*p=h[2];else*p=h[3];p++;}
if(i+1==r){h="7654";m--;e--;}E(i==r){s--;}E(i>r){s--;e++;}
else{s++;e--;}p++;}for(p=f+V(r)/2-1,i=0;i<r;i++)*++p=48;
for(i=0;i<8;i++)if(i>=x&&i<y){G(64);}else G(32);
y=y==8?0:y;q:p=f+V(r)/2-1;*p='o';switch(x){
C 0:F R 45 C 1:F U R 47 C 2:F U H C 3:F U L 92
C 4:F L 45 C 5:F D L 47 C 6:F D H C 7:F D R 92;}
if(y!=8){x=y;y=8;goto q;}puts(f);}
also, the #defines look rather ugly, but they save about 200 bytes so I kept them in, anyway. It is valid ANSI C89/C90 and compiles with very few warnings (two about atoi and puts and two about crippled form of main).